\newproblem{lay:5_1_25}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 5.1.25}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Let $\lambda$ be an eigenvalue of an invertible matrix $A$. Show that $\lambda^{-1}$ is an eigenvalue of $A^{-1}$. [\textit{Hint}: suppose a non-zero $\mathbf{x}$ satisfies
	$A\mathbf{x}=\lambda\mathbf{x}$.]
}{
  % Solution
	Suppose
	\begin{center}
		$A\mathbf{x}=\lambda\mathbf{x}$
	\end{center}
	Let's multiply on both sides by $A^{-1}$
	\begin{center}
		$\mathbf{x}=\lambda A^{-1}\mathbf{x}$ \\
		$\lambda^{-1}\mathbf{x}=A^{-1}\mathbf{x}$ \\
	\end{center}
	So $\lambda^{-1}$ is an eigenvalue of $A^{-1}$ and $\mathbf{x}$ is its associated eigenvector. It is noteworthy to see that $\mathbf{x}$ is an eigenvector of $A$ and of $A^{-1}$.
}
\useproblem{lay:5_1_25}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
